Chapter 30 Sources of the magnetic fieldForce

Equation

PointObjectForce

PointObjectField

DifferentialField

Is dB radial?Does dB have 1/r2 dependence?

Biot-Savart Law – Set-Up

• The magnetic field isdB at some point P

• The length element isds

• The wire is carrying asteady current of I

Cross product review

Right Hand Rule

Biot-Savart Law

For a long Wire:

ds

Permeability of free space

Magnetic field of a long wire

Magnetic field due to a straight wire

Compare with:

B for a Curved Wire Segment

• Find the field at point Odue to the wire segment

• I and R are constants

will be in radians

Magnetic field due to a current loop

The perpendicular componentscancel by symmetry.

Magnetic field due to a current loopAt the center:

Magnetic dipole moment vectorfor a single loop:

When x>>R:

Comparison to an electric dipole

+

-q

q

Biot-Savart Law:

5. Determine the magnetic field at a point Plocated a distance x from the corner of aninfinitely long wire bent at a right angle, asshown in Figure P30.5. The wire carries asteady current I.

7. The segment of wire in Figure P30.7carries a current of I = 5.00 A, where theradius of the circular arc is R = 3.00 cm.Determine the magnitude and direction ofthe magnetic field at the origin.

10. A very long straight wire carries currentI. In the middle of the wire a right-anglebend is made. The bend forms an arc of acircle of radius r, as shown in FigureP30.10. Determine the magnetic field at thecenter of the arc.

Force between two parallel wires

a

Force between two parallel wires

If the currents are in the samedirection, the force is attractive.

If the currents are in the oppositedirection, the force is repulsive.

Historical definition of the Ampere

Historical definition of the Coulomb

If 2 long parallel wires 1.0 m apart have the samecurrent in them and the force per unit length on eachwire is 2.0 x 10-7 N/m, the current is 1.0 Ampere

If the current is 1.0 Ampere, then 1.0 Coulomb is theamount of charge passing through a cross section in1 second.

16. Two long, parallel conductors, separated by 10.0 cm, carry currents in the samedirection. The first wire carries current I1 = 5.00 A and the second carries I2 = 8.00A. (a) What is the magnitude of the magnetic field created by I1 at the location ofI2? (b) What is the force per unit length exerted by I1 on I2? (c) What is themagnitude of the magnetic field created by I2 at the location of I1? (d) What is theforce per length exerted by I2 on I1?

18. Two long, parallel wires are attracted to each other by a force per unit length of 320μN/m when they are separated by a vertical distance of 0.500 m. The current in theupper wire is 20.0 A to the right. Determine the location of the line in the plane of thetwo wires along which the total magnetic field is zero.

63. Two long, parallel conductors carry currents in the samedirection as shown in Figure P30.63. Conductor A carries acurrent of 150 A and is held firmly in position. Conductor Bcarries a current IB and is allowed to slide freely up anddown (parallel to A) between a set of nonconducting guides.If the mass per unit length of conductor B is 0.100 g/cm,what value of current IB will result in equilibrium when thedistance between the two conductors is 2.50 cm?

Introduction to Ampere’s Law

Recall the definition of electric potential:

What is the value of the integral over a closed path for anyelectric field?

Let’s try the same thing for a magnetic fieldaround a current carrying wire.

Ampere’s Law

• The integral is around any closed path• The current is that passing through the surface bounded by

the path• Like Gauss’s Law, useful in finding fields for highly

symmetric problems

This result has been shownexperimentally to be true in general

Applying Ampere’s Law

• Select a surface– Try to imagine a surface where

the electric field is constanteverywhere. This isaccomplished if the surface isequidistant from the charge.

– Try to find a surface such thatthe electric field and thenormal to the surface are eitherperpendicular or parallel.

• Determine the charge insidethe surface

• If necessary, break theintegral up into pieces andsum the results.

• Select a path– Try to imagine a path where the

magnetic field is constanteverywhere. This is accomplishedif the surface is equidistant fromthe charge.

– Try to find a path such that themagnetic field and the path areeither perpendicular or parallel.

• Determine the current insidethe surface

• If necessary, break theintegral up into pieces andsum the results.

Example: Magnetic field inside a wire

Example: Solenoid

Example: Solenoid

0 0

small

Example: Toroid

Inside:

Outside:

21. Four long, parallel conductors carry equal currents of I =5.00 A. Figure P30.21 is an end view of the conductors. Thecurrent direction is into the page at points A and B (indicatedby the crosses) and out of the page at C and D (indicated bythe dots). Calculate the magnitude and direction of themagnetic field at point P, located at the center of the squareof edge length 0.200 m.

29. A long cylindrical conductor of radius R carries acurrent I as shown in Figure P30.29. The current densityJ, however, is not uniform over the cross section of theconductor but is a function of the radius according to J =br, where b is a constant. Find an expression for themagnetic field B (a) at a distance r1 < R and (b) at adistance r2 > R, measured from the axis.

24. The magnetic field 40.0 cm away from a long straight wire carrying current 2.00A is 1.00 μT. (a) At what distance is it 0.100 μT? (b) What If? At one instant, the twoconductors in a long household extension cord carry equal 2.00-A currents inopposite directions. The two wires are 3.00 mm apart. Find the magnetic field 40.0cm away from the middle of the straight cord, in the plane of the two wires. (c) Atwhat distance is it one tenth as large? (d) The center wire in a coaxial cable carriescurrent 2.00 A in one direction and the sheath around it carries current 2.00 A in theopposite direction. What magnetic field does the cable create at points outside?

Magnetic Flux

• The magnetic field in this element is B• dA is a vector that is perpendicular to the surface• dA has a magnitude equal to the area dA• The magnetic flux B is

• The unit of magnetic flux is T.m2 = Wb– Wb is a weber

Gauss’ Law in Magnetism

• Magnetic fields do not begin or end at anypoint– The number of lines entering a surface equals

the number of lines leaving the surface• Gauss’ law in magnetism says:

Displacement Current• Ampere’s law in the original form

is valid only if any electric fieldspresent are constant in time

• Maxwell added an additional termwhich includes a factor called thedisplacement current, Id

• The displacement current is not thecurrent in the conductor– Conduction current will be used to

refer to current carried by a wire orother conductor

Ampere’s Law – General Form

• Also known as the Ampere-Maxwell law

• Magnetic fields are produced both byconduction currents and by time-varyingelectric fields

Ferromagnetism

• Domains• Curie Temperature• Electron orbits align

with an externalmagnetic field